HDU 5407(CRB and Candies-OEIS)

CRB and Candies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 646    Accepted Submission(s): 321

Problem Description

CRB has N different candies. He is going to eat K candies.
He wonders how many combinations he can select.
Can you answer his question for all K (0 ≤ K ≤ N )?
CRB is too hungry to check all of your answers one by one, so he only asks least common multiple(LCM) of all answers.

 

Input

There are multiple test cases. The first line of input contains an integer T , indicating the number of test cases. For each test case there is one line containing a single integer N .
1 ≤ T ≤ 300
1 ≤ N ≤ 106

 

Output

For each test case, output a single integer – LCM modulo 1000000007( 109+7 ).

 

Sample Input

   
   
   
   

    
    
    
    5
1
2
3
4
5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
3
12
10
   
   
   
   

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

 

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在OEIS上，查出答案=lcm(1,2,..,n)/n

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
ll gcd(ll a,ll b) {
	if (b==0) return a;return gcd(b,a%b);	
}
ll lcm(ll a,ll b){
	return a/gcd(a,b)*b;
}
int n;

int p[MAXN],tot=0;
bool b[MAXN]={0};
void make_prime()
{
	int n=1000005;
	Fork(i,2,n) {
		if (!b[i]) b[i]=1,p[++tot]=i;
		
		For(j,tot) {
			if (p[j]*i>n) break;
			b[p[j]*i]=1;
			if (i%p[j]==0) break;
		}
	}
	
}

ll pow2(ll a,ll b){
	if (b==0) return 1;
	if (b==1) return a;
	ll t=pow2(a,b/2);
	t=t*t%F;
	if (b&1) t=t*a%F;
	return t;
}

int main()
{
//	freopen("B.in","r",stdin);
	
	make_prime();
//	For(i,100) cout<<p[i]<<' ';
	
	int T; cin>>T;
	while(T--) {
		cin>>n; ++n;
		ll ans=1;
		
		ll c=1;
	//	For(i,n/2) c=c*(n-i+1)/i,ans=lcm(ans,c),cout<<c<<endl;
		
		For(i,tot) {
			if (p[i]>n) break;
			
			double k=log(n)/log(p[i]);
			int k2=(int)k;
			
			
			ans=mul(ans,pow2(p[i],k2));
			
		}
		ans=mul(ans,pow2(n,F-2));
		
		cout<<ans<<endl;
	} 	
	
	return 0;
}